YES
O(n^5)
TRS:
 {b(b(a(x))) -> a(b(b(x))),
     a(a(x)) -> b(b(x))}
 DUP: We consider a non-duplicating system.
  Trs:
   {b(b(a(x))) -> a(b(b(x))),
       a(a(x)) -> b(b(x))}
  Matrix Interpretation:
   Interpretation class: triangular
       [X4]    [1 1 0 0 0][X4]   [1]
       [X3]    [0 0 1 1 1][X3]   [0]
   [a]([X2]) = [0 0 1 0 0][X2] + [1]
       [X1]    [0 0 0 0 0][X1]   [0]
       [X0]    [0 0 0 0 0][X0]   [0]
   
       [X4]    [1 1 0 0 0][X4]   [0]
       [X3]    [0 0 1 0 0][X3]   [0]
   [b]([X2]) = [0 0 1 0 0][X2] + [0]
       [X1]    [0 0 0 0 0][X1]   [0]
       [X0]    [0 0 0 0 0][X0]   [0]
   
   
   Qed